#Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length. 
#
# Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory. 
#
# Example 1: 
#
# 
#Given nums = [1,1,2],
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#Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.
#
#It doesn't matter what you leave beyond the returned length. 
#
# Example 2: 
#
# 
#Given nums = [0,0,1,1,1,2,2,3,3,4],
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#Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively.
#
#It doesn't matter what values are set beyond the returned length.
# 
#
# Clarification: 
#
# Confused why the returned value is an integer but your answer is an array? 
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# Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well. 
#
# Internally you can think of this: 
#
# 
#// nums is passed in by reference. (i.e., without making a copy)
#int len = removeDuplicates(nums);
#
#// any modification to nums in your function would be known by the caller.
#// using the length returned by your function, it prints the first len elements.
#for (int i = 0; i < len; i++) {
#    print(nums[i]);
#} Related Topics Array Two Pointers



#leetcode submit region begin(Prohibit modification and deletion)
class Solution(object):
    def removeDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums:
            return 0
        i = 0
        for j in range(1, len(nums)):
            if nums[i] != nums[j]:
                i += 1
                nums[i] = nums[j]
        return i+1
#leetcode submit region end(Prohibit modification and deletion)
